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Is there some known way to create the Mandelbrot set (the boundary), with an iterated function system (IFS)?

Julia sets can be formed by iterating the two functions $z \mapsto \pm \sqrt{z-c},$ and this quickly converges to the Julia set. Formally, the two functions form a Hutchinson operator, and the Julia set is the unique compact fixed set. That is, $J = f_1(J) \cup f_2(J)$ where $f_1,f_2$ are the two functions, and $J$ is the Julia set.

However, Mandelbrot is the iteration of many functions, with start value 0, so there is no obvious way to define an IFS. Is there some non-obvious way?

That is, find maps $g_1,\dots,g_n$ s.t. the Mandelbrot $M$ can be expressed as $M = \bigcup_i g_i(M),$ where each $g_i$ maps $M$ to some subset of $M,$ but not the entire $M.$

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  • $\begingroup$ What do you mean by "create"? If $J_c$ is the Julia set corresponding to $c\in \mathbb C$, let $X_{c,n}$ be the set you create by taking the $n$-th iterate of your 2-valued function $z\longmapsto \sqrt{z−c}$, with the initial iterate being $z_0=0$. Do you want "create" to mean that $\cap_{n=1}^\infty \overline{\cup_{k=n}^\infty X_{c,k}} = J_c$ or something like that? $\endgroup$
    – Ryan Budney
    Commented Sep 3, 2011 at 21:26
  • $\begingroup$ Yes, something like that. I mean, the Julia set is the fixed set for a certain Hutchinson operator, en.wikipedia.org/wiki/Hutchinson_operator with the two functions given above. $\endgroup$
    – Per Alexandersson
    Commented Sep 4, 2011 at 7:27
  • $\begingroup$ Look at Numberphile concerning the Mandelbrot Set. It's a YouTube Channel. $\endgroup$
    – Mr Pie
    Commented Oct 7, 2017 at 21:54

2 Answers 2

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I know very little about the Mandelbrot set or complex dynamics, but there are several features which strongly suggest that its boundary is not the attractor of an iterated function system, at least a reasonable one.

For example.

  • Around Misiurewicz points (which are only countable but dense), the Mandelbrot set looks locally like the corresponding Julia set, in particular it looks locally very different for each Misiurewicz point. Under some natural assumptions, sets invariant under IFS's look locally like the set itself.
  • If an attractor of an IFS is connected, it is locally connected. The boundary of the Mandelbrot set is connected but it is a famous open problem whether it is locally connected.
  • Under some weak contractivity assumptions on the generating maps, attractors of IFS's cannot have full dimension in the ambient space, while the boundary of the Mandelbrot set has dimension 2.

Also it is not quite true that all Julia sets are fixed sets for Hutchinson operators, at least under the usual definition of Hutchinson operator which requires strict contractivity of the maps (the maps $z\to\pm\sqrt{z-c}$ are in general not contractions, even on the Julia set itself)

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  • $\begingroup$ Oh, yes it is required that the functions are contractive. However, I suspect that under certain milder assumptions, the attractor is still unique. (Numerical experiments suggests this, for example, the software Apophysis among other happily produces Julia sets from the maps above, even though there is no proof that I know of that says that it works). A set of functions that gives the Mandelbrot set suffices to me, even thought there is no proof that this really is the case. $\endgroup$
    – Per Alexandersson
    Commented Sep 5, 2011 at 11:26
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You may take a look at Barnsley, Fractals everywhere, pp. 318f, Figures viii.237 and viii.238 (http://books.google.de/books?id=oh7NoePgmOIC&lpg=PP1&hl=de&pg=PA319#v=onepage&q&f=false). It looks not exactly like the Mandelbrot set, but there is a similarity. Unfortunately Im not able to reproduce these images (using algorithm 2.1 from page 302).

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